Boomerang Generation - Short Term: In a 2010 Pew Research Center survey, suppose

Question

Boomerang Generation - Short Term: In a 2010 Pew Research Center survey, suppose 83 out of 396 randomly selected young adults (ages 18–34) had to move back in with their parents after living alone. In a 2012 survey, suppose 199 out of 803 young adults had to move back in with their parents. The table below summarizes this information. The standard error (SE) is given to save calculation time if you are not using software.

Data Summary:

SE = 0.02604

The Test: Test the claim that a greater proportion of all young adults moved back in with their parents in 2012 than in 2010. Use a 0.01 significance level.

(a) Letting p1 be the proportion of young adults that had to move back in with their parents in 2012 and p2 be the proportion from 2010, calculate the test statistic using software or the formulaz =

where p is the hypothesized difference in proportions from the null hypothesis and the standard error (SE) given with the data. Round your answer to 2 decimal places.
z = 1
To account for hand calculations -vs- software, your answer must be within 0.01 of the true answer.

(b) Use software or the z-table to get the P-value of the test statistic. Round to 4 decimal places.
P-value = 2

(c) What is the conclusion regarding the null hypothesis?

reject H0

fail to reject H0    

(d) Choose the appropriate concluding statement.

The data supports the claim that a greater proportion of all young adults moved back in with their parents in 2012 than in 2010.

There is not enough data to support the claim that a greater proportion of all young adults moved back in with their parents in 2012 than in 2010.     We have proven that a greater proportion of all young adults moved back in with their parents in 2012 than in 2010.

We reject the claim that a greater proportion of all young adults moved back in with their parents in 2012 than in 2010.

number who total number Proportion Year moved back (x) in survey (n) p = (x/n) 2012 199 803 0.24782 2010 83 396 0.20960

Explanation / Answer

(a) Letting p1 be the proportion of young adults that had to move back in with their parents in 2012 and p2 be the proportion from 2010, calculate the test statistic using software or the formulaz

so Test statistic

Z= (p1 p2) p / SE = (0.24782 - 0.20960)/ 0.02604 = 1.47

(b) P - value = Pr (Z >1.47) = 0.0708

(c) We fail to reject the null hypothesis as Z < Zcr

(d) Conclusion : We reject the claim that a greater proportion of all young adults moved back in with their parents in 2012 than in 2010. Option D is correct.

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